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Topic: Fermat revised ?
Replies: 8   Last Post: Aug 18, 2007 3:51 PM

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stephane.frion@gmail.com

Posts: 14
Registered: 8/15/07
Re: Fermat revised ?
Posted: Aug 15, 2007 11:47 PM
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On Aug 15, 8:46 pm, stephane.fr...@gmail.com wrote:
> On Aug 15, 6:23 pm, drmwec...@gmail.com wrote:
>
>
>
>
>

> > For starters, yes, there are solutions.
>
> > The first I found is for n = 2, with x = 27, y = 18, z = 9.
> > For n = 3, try x = 256, y = 64, z = 32.

>
> > Just playing around with QuickBASIC and a calculator program here...
>
> > Best, Mike
>
> > Dr. Michael W. Ecker
> > Associate Professor of Mathematics
> > Pennsylvania State University
> > Wilkes-Barre Campus
> > Lehman, PA 18627

>
> > On Aug 15, 8:49 pm, stephane.fr...@gmail.com wrote:
>
> > > I consider the equation :
> > > X^n + Y^(n+1) = Z^(n+2)
> > > Where X, Y, Z are integer non-null and n integer > 1
> > > Is there a solution to this equation?

>
> > > With my pool skill in math, I looked first at X^n + Y^(n+1) = (X^2)^n,
> > > chosing Z =X . But I still can see anything

>
> > > I saw somewhere that if n = 2, then there is solution if X, Y and Z
> > > are relatively prime . What about n = 3 ?

>
> > > SF
> > > ---------------------------------------------------------------------------???-------
> > > " I'm not native english speaker so apologize if you can't understand
> > > me "- Hide quoted text -

>
> > - Show quoted text -
>
> I think I understand .
> To start a particular case, as there is no constraint on x, y and z ,
> I can suppose that
> x=b*z and y=a*z with a and b integers
> Then z^(n+2)=y^(n+1) + x^n <=> z^(n+2) = a^(n+1)*z^(n+1) + b^n * z^n
> <=> z^2 = a^(n+1)*z + b^n as z diff 0
> There is solution in real if D=a^2(n+1)+4*b^n >=0, in addition this
> need to be a perfect square
> The strong condition in this particular case will be to have solution
> will be to have -a^(n+1)+D^0.5 or -a^(n+1)-D^0.5 even number
>
> in your case n=2 , a=2 , b=3 then D=100 then the solutions are 9 or -1
> which provide
> 9 => x = 27 , y = 18 , z = 9
> -1 => x = -3, y = -2 , z=-3
>
> I wonder if there is an easy way to find a, b such as -a^(n+1)+D^0.5
> or -a^(n+1)-D^0.5 is an even number
>
> SF- Hide quoted text -
>
> - Show quoted text -


Typo error ... -1 => x = -3, y = -2 , z=-1 corrected
>




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